# Mechanics of solids question from my mid semester exam

1. Apr 21, 2010

### ssar

1. The problem statement, all variables and given/known data
A small dam of triangular shape as shown is made from concrete. Find the stresses on the foundation at A and B. Assume concrete and water weighs 25kN/m^3 and 10kN/m^3
(All dimensions in the attached picture are in metres)

2. Relevant equations
sigma = force/cross sectional area
bending stress = M.y/I (I'm thinking we might need to use this)
centroid of triangle is a 1/3 from the high side

3. The attempt at a solution

I'm pretty much all over the place as I have no direction. So pretty much all that I've jotted down as just been ideas that lead nowhere.

I labeled the top of the dam as C

Area ABC = 6m^2

One idea I was told was to find the bending stress at point A and B, take I to be value of the dam, find the Max M of the dam and the value of centroids to be 1m for point B and 2m for point A.

I'm not too sure if my FBD is correct because I think it may be a fixed support but this is what my friend thinks it will look like. See below.

I just need the right direction or the right idea. I'm thinking that I have to try and convert those weights into a force, just not sure how to work out the force for the water.

2. Apr 21, 2010

### ponjavic

The force from the water per m is:
The integral of the pressure due to the water (from 0 to h)

3. Apr 23, 2010

### pongo38

Consider a 1 m length of wall (perpendicular to the page). Call it, say AA'BB'. The section you need to consider stresses on is that plan section AA'BB'. Then apply the equation for combined axial and bending stresses N/A +- M/Z

The Ma in your diagram. What value do you think it has?

4. Jun 16, 2010

### ssar

Ok well I know it has been a while but we just had a tutorial on this topic and did a similar question so heres my attempt.

Here is the question

http://yfrog.com/0fquestionzfj

Wc = weight of concrete
Ww = Weight of water
Pc = force of concrete
Pw = force of water
Fw = stress of water (i think)

Wc = 25kN/m^3

Density of water = 1000kg/m^3
Ww = 1000x9.81
= 9.81kN/m^3

Fw = 9.81 x 3.3 = 32.373 kN/m^2

Then since its a triangle, Pc acts 1/3 from the high side of 2.16m and assuming 1m wall perpendicular to the page.

Pc = 0.5*2.16*3.6*1*25 = 97.2 kN
Pw = 0.5*32.373*3.3*1 = 53.415 kN (in this line I was a bit confused as to why its you multiply it by 0.5)

Then consider cross section of the base where it looks like this

using Z = I/c

Z = (1/12*1*2.16^3)/1.08
= 0.7776m^3

SigmaA = -97.2/2.16 + 97.2*0.36/0.7776 - 53.415*1.1/0.7776
SigmaA = -75.56 kN/m^2

Similarly for SigmaB

SigmaB = -14.44 kN/m^2

I think the only value that I'm not sure about is my Fw/Pw. But how does that look?

5. Jun 17, 2010

### pongo38

This looks ok to me. In your comment "(in this line I was a bit confused as to why its you multiply it by 0.5)" the pressure distribution from the water is triangular with the maximum 32,373 at the bottom. So the factor of 0.5 is taking the AVERAGE pressure.

6. Jun 17, 2010

### ssar

Oh ok, I kept confusing myself with it being a triangle or just being a rectangle. Thanks for the clarification.